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This one is very simple but I guess I am missing out something and hence finding it hard to understand.

When a coupling capacitor is connected in an amplifier circuit, one of its plates is connected to the DC voltage coming out of a voltage divider. This tells me that it has a voltage on it. So now when the electrons from the AC signal come, won’t the electrons present on the DC side plate repel them? How does this AC signal cause electrons on the DC side to repel and move towards the transistor's input? I know that electric forces act on charges such that the force felt by two interacting charges has the same magnitude.

I don't know if I have framed my question correctly, but for me it’s simply not easy to understand that we basically charge the capacitor from both sides.

I feel I am missing something vital.

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  • \$\begingroup\$ Capacitors may be analyzed from an electrostatic view to a time domain view with DC to AC characteristics Ic=CdV/dt then also as an AC impedance in frequency domain where Z is inverse to f. Don't try to confuse yourself with electrostatics and AC analysis \$\endgroup\$ Jun 1, 2019 at 13:20
  • \$\begingroup\$ ibiblio.org/kuphaldt/socratic/model/mod_capacitor.pdf then follow the Socratic reasoning and jump right into Python \$\endgroup\$ Jun 1, 2019 at 13:21
  • \$\begingroup\$ Shai, be careful with grammar and stuff, as far as I can see, you write English well, but you most likely wrote your question in a hurry. Still, you should minimise these. \$\endgroup\$ Jun 1, 2019 at 13:53
  • \$\begingroup\$ Other than that, welcome to Electrical Engineering Stack Exchange! You can upvote any answer you feel like helped you with your question and you can also accept only one, if it solved your doubts 100%. \$\endgroup\$ Jun 1, 2019 at 16:57

2 Answers 2

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. OP's circuit.

So now when the electrons from the ac signal comes, won’t the electrons present on the dc side plate repel them.

You are forgetting a couple of things:

  1. For current to enter C1 on the left side then the same current must leave on the right side returning to the V1 source via R2 (or via R1 and V2). That's the nature of circuit theory. V1 is pushing charge around the circuit but it is a circuit.
  2. Not that it matters terribly, but the right side of C1 has positive charge on it so according to your theory negative charge should be attracted to the left side of C1.
  3. V1 has voltage or "electro-motive force". It can push charge around a circuit.

Notice here that I am deliberately using the word charge as current is not always electron flow. This also allows you to think in terms of conventional current flow and will save you much confusion.

So what is happening then? What has helped me in understanding circuit theory is to thing of capacitors as having a voltage across them (which could be zero) and in the short term the capacitor keeps the voltage across it constant. In Figure 1 C1 is charged to V2/2. Now as V1 voltage increases the left side of C1 is "lifted" and the right side lifts by the same amount but immediately starts to discharge through the combination of R1 and R2 back to the V2/2 voltage. You should be able to follow this through and see what happens when V1 decreases and goes negative.

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  • \$\begingroup\$ Thanks a lot. I understood your explanation. I read couple of theories related to capacitors recently and got to know how exactly they work (basically how charges on the capacitor's one plate affect the voltage levels on the opposite plate) and then I came back to this. If V1 goes negative then capacitors voltage will drop below V2/2 and it will start charging back to V2/2 if I am not wrong. \$\endgroup\$ Mar 4, 2020 at 17:53
  • \$\begingroup\$ I think you've got it. Don't forget to "accept" one of the answers if it gives you the solution you required. You can also upvote other answers that were useful. \$\endgroup\$
    – Transistor
    Mar 4, 2020 at 18:30
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I agree with @Sunnyskyguy EE75 in that you are mixing two things: circuit theory and electrostatics.

The coupling capacitor's purpose is to block any DC coming from another stage which could upset the DC working conditions of the stage we are currently thinking about. It is supposed to block any DC coming at the input, frankly.

As I said in the first line, I don't think it is too useful for your understanding to think about the force present between charges due to electric interactions. Yet, since you are mainly mentioning it, I believe I can explain it that way.

You may have two charges (electrons) posessing the same sign, but that does not mean they will stop each other and (eventually?) go in opposite directions. The DC current is created by an ordered electron movement, caused by the DC voltage source, which provides an Electro-Motive Force, causing the charges to react as mentioned. The two sources keep pushing the electrons, as @Transistor said. So they will repel each other as they move in one direction.

If you take two magnets and make them meet, by throwing them at each other, such that the ends with the same polarity meet, they will stop each other/ maybe move back in opposite directions. Now if you keep pushing one of the magnets towards the other, they will move in the direction you are pushing while still repelling each other. The force between charges is a concept from electrostatics. The electrons experiencing the same force (in modulus) but with opposite orientations is a case occurring when there are no other external forces applied.

If you now think in terms of circuit theory and in terms of voltage, the DC signal will be superimposed on the AC input (as DC offset). This means that every point from the AC wave will be summed with the DC voltage value. So instead of having AC between -9 V to 9 V, you will get a signal alternating between -7 V and 11 V, if you apply a DC signal of 2 V. Of course, in the field of amplifiers, you will be dealing with weak input signals, such as something with \$V_{peak}=10 mV\$.

I think you might find it helpful to first analyse your question from a circuit theory perspective. Then, after you understand it well enough, you can move onto explaining with the help of electrostatics. I think it's useful to start simple when you try to learn something new.

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